Question

please answer this question ​

Answer 1

Answer:

Let first term= a common difference = d

16th term = a + 15d

3rd term = a+ 2d

Given

a+ 15 d = 5( a + 2d)

a + 15d = 5a + 10d

5 d = 4a

so [a = 5/4 d ]

Now 10th term = a + 9d = 41

Putting value of a in this

5/4 d + 9d = 41

[ 5d + 36d ] = 41 * 4

41 d = 41 *4

d = [41* 4] / 41

d = 4

a = 5/4 * 4

a= 5

Sum of first 15 terms = 15/2 [ 2a + 14d]

= 15/2 [ 10 + 56 ]

= 15/2 [ 66]

= 15 [ 33]

= 495

Was this answer helpful

Answer 2

Answer:

i all so need this question answer

Step-by-step explanation:

can you explain it